Question 1141514: Need help with the following question. Please show work. Much appreciated, thank you!
It is known that 20% of drivers in Canada will drive after smoking marijuana. If 18 drivers are randomly selected, what is the probability that:
a) 4 or more will drive after smoking marijuana?
b) At least one will drive after smoking marijuana?
c)Less than 3 will drive after smoking marijuana?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! Use the binomial probability formula
:
Probability (P) (k successes out of n trials) = nCk * p^k * (1-p)^(n-k), p is the probability of success, nCk = n!/(k! * (n-k)!)
:
for this problem, p = 0.20, n=18
:
(a) summation from i = 4 to 18 of P(i out of 18 trials) = 0.4999
:
(b) 1 - P(0 successes out of 18 trials) = 1 - 0.0180 = 0.9820
:
(c) summation from i = 0 to 2 of P(i out of 18 trials) = 0.2713
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